Question

$\int \frac{\left[f\right(x\left)g^{{}^{\prime }}\right(x)+g(x\left)f^{{}^{\prime }}\right(x\left)\right]}{f\left(x\right).g\left(x\right)}\left[logf\right(x)+logg(x\left)\right]dx$ is equal to

• A. $f\left(x\right)g\left(x\right)log\left(f\right(x\left)g\right(x)+C$
• B. $\frac{1}{2}\left[logf\right(x\left)g\right(x){]}^2+C$
• C. $\left[logf\right(x\left)g\right(x){]}^2+C$
• D. $logf\left(x\right).g\left(x\right)+C$

Answer 1

The correct option is B $\frac{1}{2}\left[logf\right(x\left)g\right(x){]}^2+C$

$\int \frac{\left[f\right(x\left)g^{{}^{\prime }}\right(x)+g(x\left)f^{{}^{\prime }}\right(x\left)\right]}{f\left(x\right).g\left(x\right)}\left[logf\right(x)+logg(x\left)\right]dx=\int \frac{\left[f\right(x\left)g^{{}^{\prime }}\right(x)+g(x\left)f^{{}^{\prime }}\right(x\left)\right]}{f\left(x\right).g\left(x\right)}\left[logf\right(x\left)g\right(x\left)\right]dx=\int \frac{logt}{t}dt\left[puttingf\right(x).g(x)=t\Rightarrow [f\left(x\right)g^{{}^{\prime }}x+g\left(x\right)f^{{}^{\prime }}\left(x\right)dx=dt]=\frac{1}{2}(logt{)}^2+C=\frac{1}{2}\left[logf\right(x).g(x){]}^2+C$